Question

The total number of α and β particles emitted in the nuclear reaction $_{92}^{238}U\to _{82}^{214}Pb$ is:

Answer 1

Hint: Alpha particles (α) is a helium nucleus. The α decay is shown as $_{Z}^{A}X\xrightarrow{\alpha }_{Z-2}^{A-4}Y$, where 'A' is the atomic mass and 'Z' is the atomic number and β decay is shown as $_{Z}^{A}X\xrightarrow{\beta }_{Z+1}^{A}Y$ . To find the alpha particle subtract the atomic masses of X and Y and divide it with 4 and for beta particles use the formula Atomic no of X= Atomic no. of Y + 2α-β.

Complete answer:
From your chemistry lessons you have learned about the radioactivity, α- decay and β-decay.
Alpha decay (α-decay)
It is a type of decay in which a nucleus transforms itself into a different nucleus by emission of alpha particles. Nucleus of helium $(_{2}^{4}He)$ is the alpha particle. We know that helium contains two neutrons and two protons. Therefore after the emission the atomic mass of the nucleus which is going through emission reduces to 4 and atomic no. reduces to 2,
Hence the change of $_{Z}^{A}X$ to the emitting nucleus $_{Z-2}^{A-4}X$ is expressed as ,
\[_{Z}^{A}X\to _{Z-2}^{A-4}Y+_{2}^{4}He\]
Beta decay (β decay)
Beta decay is a type of decay in which a nucleus decay spontaneously by the emission of only electrons or positrons. It is of two types beta plus decay and beta minus decay.
In beta plus decay $({{\beta }^{+}})$ there is the emission of positron (positive electron) for example,
\[_{Z}^{A}X\to _{Z-1}^{A}Y+{{e}^{+}}+v\]
In beta minus decay $({{\beta }^{-}})$ there is the emission of electron from the nucleus for example, \[_{Z}^{A}X\to _{Z+1}^{A}Y+{{e}^{-}}+{{v}^{-}}\]
In the question we have to find the alpha and beta particle in $_{92}^{238}U\to _{82}^{214}Pb$
It is also written as
$_{92}^{238}U\xrightarrow{\alpha ,\beta }_{82}^{214}Pb+_{2}^{4}He+_{-1}^{0}e$
So, to find the alpha particle you can solve it by Atomic mass of Pb = Atomic mass of U - 4α ……..(1)
Atomic mass of U = 238
Atomic mass of Pb = 214
So by putting the values in equation(1), you will get
206 = 238 - 4α
\[\therefore \alpha =\dfrac{238-214}{4}=6\]
Now find the beta particle by using formula Atomic no of U= Atomic no. of Pb + 2α-β. Where atomic no. of U is 92 and atomic no. of Pb is 82 and the value of α that we have find is 6
By putting the values you will get,
92 = 82 + 2×6 – β
Therefore, β = 92- 94 = 2

Thus the no. of alpha and beta particles is 6 and 2.

Note:
The reaction that is given about is not a direct decay because it has gone under many alpha and beta decay. Positron does not refer to protons, it is a positively charged electron and stands for neutrino , it is a neutral particle with no mass or little mass. In alpha decay there is no requirement of external energy. The decrease in atomic mass is only caused by alpha decay and not by the beta decay ( it can only increase or decrease atomic number).